3.14 Sspd_chapter1_part 8continued_electron in an infinite potential

SSPD_Chapter 1_Part8 continued_ SOLUTION OF SCHRODINGER EQUATION FOR AN ELECTRON IN AN INFINITE POTENTIAL WELL.

Fig(1.24) Electron in 1-D infinite potential well. Probability Amplitude ψ(z,t) and probability density |ψ(z,t)|^ 2 are respectively plotted along the spatial axis-z axis.

Writing the Schrodinger Equation [ Appendix XXIX] for an electron in an infinite potential well where V(z)=0 we get:

(ћ 2 /2m) ∂ 2 ψ /∂z 2 + E ψ = 0 1.48

∂ 2 ψ /∂z 2 + (2mE/ћ 2 )ψ = 0

According to Linear Operator Theory, second order linear differential equation has two solutions: the complementary solution also known as transient solution or the natural solution and the second solution is the particular solution also known as the steady state solution or the forced solution.

The total solution = transient solution + steady state solution.

The transient solution depends on the characteristic equation of the system. A second order linear system has second degree characteristic equation which has two roots and two arbitrary constants which need two boundary conditions.

The steady state solution depends on the forcing function or the driving function. If the forcing function is a harmonic function then the steady state solution is a sinusoidal solution. If the forcing function is a constant the steady state output is also a constant.

In Eq.(1.48) Right Hand Side is zero hence steady state solution is zero.

The characteristic equation is:

D ² + 2mE/ћ ² = 0

Where D is the differential operator.

The roots are imaginary:

D ₁ = -i[√(2mE)]/ ћ and D ₂ = +i[√(2mE)]/ ћ

Therefore the solution is harmonic. If the roots were real then the solution would be hyperbolic or exponential.

The solution of Eq.(1.48) is:

ψ = A.Exp[+i{(√(2mE))/ ћ}x] + B.Exp[-i{(√(2mE))/ ћ}x] ;

From two boundary conditions the two arbitrary constants are determined.

At x = 0, ψ(0) = 0 and at x = W ( the width of the potential well), ψ(W) = 0.

From these two boundary conditions we obtain two simultaneous equations.

A + B=0;

A.Exp[+i{(√(2mE))/ ћ}W] + B.Exp[-i{(√(2mE))/ ћ}W]= 0;

The determinant is:

Δ = 1 1

Exp[+i{(√(2mE))/ ћ}W] Exp[-i{(√(2mE))/ ћ}W]

Therefore Δ = Exp[-i{(√(2mE))/ ћ}W] - Exp[+i{(√(2mE))/ ћ}W]

If these two simultaneous equations are to have non-trivial solutions then the determinant of the set of simultaneous equations should be zero.

Therefore Δ = Exp[-i{(√(2mE))/ ћ}W] - Exp[+i{(√(2mE))/ ћ}W]=0

Therefore Exp[-i{(√(2mE))/ ћ}W] = Exp[+i{(√(2mE))/ ћ}W]

Therefore Cos[{(√(2mE))/ ћ}W] –i Sin[{(√(2mE))/ ћ}W]

= Cos[{(√(2mE))/ ћ}W] +i Sin[{(√(2mE))/ ћ}W]

Or –i Sin[{(√(2mE))/ ћ}W] = i Sin[{(√(2mE))/ ћ}W]

This relationship is satisfied only when Sin[{(√(2mE))/ ћ}W]=0;